I would like to see if anyone can comment on these two opposing points argued below. The first argument denotedby double arrows `>>` states that a shorter wavelengthcan have its frequency adjusted to equal the frequencyof a slightly longer wavelength simply by slowingdown the speed of the shorter wavelength. Implicit inthis argument is the fact that it is possible to reduce the observed frequency of a wavelength by reducing its speed.

The second argument at the bottom denoted by single `>`claims that this is not true and that it is impossible to adjust the speed of a shorter wavelength so that its observed frequency matches that observed from a slightly longer wavelength. I would like to see if anyone can say which argument is correct.

>> Here look at this example.>> lets say we have the shorter wavelength as .8 meter>> and the longer wavelength at 1 meters>> so we then get this visual below. A is longer>> than B, X is the observor point. Direction>> of movement is r-l>>>> X meters>> A 0 1 2 3 4 <<>> a b c d e>>>> B 0 .8 1.6 2.4 3.2 4 <<>> f g h i j k>>>> If they both travel at the same speed lets>> say 1 meter a second then they are not at the same>> frequency as B is observed to have 5 peaks per sec>> and A is observed to have 4 per peaks per second.>> However if B travels at .8 meters a second then>> they will both have the same frequency observed at>> X. So now, every second, 1 peak from A passes X and>> every second 1 peak from B passes X.>> So, where A`s speed is 1 m/s and B`s speed>> is .8 m/s, we get the following using the letters a-k>> under each wave denoted above,...>>>> at 0 seconds a and f pass X>> at 1 second b and g pass X>> at 2 seconds c and h pass X>> at 3 seconds d and i pass X etc etc...>> Both wavelengths now have the same frequency!>> You are wrong to say this is not possible.

>Let me try a sketch too. 'a' is the series>of peaks at frquency A, is those for B. The>horizontal scale is time, not distance.

>In the above, D is the distance from the source,>va is the speed for frequency A and vb is the>speed for frequency B.

>I have drawn in three of the delays for A and B>in the bottom diagram and you could add them>for each peak but it would get too cluttered>in this post. Notice how the delay has>aligned one pair of peaks that would otherwise>have missed but thereafter they diverge again.